Match List I with List II

Choose the correct answer from the options given below:

We first recall the characteristic transformations named after Finkelstein, Swarts, Hunsdiecker and Sandmeyer reactions.

Case A - Finkelstein reaction
Definition: Exchange of $$Cl^-$$ or $$Br^-$$ in an alkyl halide by $$I^-$$ using dry $$NaI$$ in acetone.
Result: Formation of an alkyl iodide $$R\!-\!I$$ (List II-III).

Case B - Swarts reaction
Definition: Replacement of $$Cl$$ or $$Br$$ in an alkyl halide by $$F$$ with metallic fluorides such as $$AgF,\; Hg_2F_2,\; SbF_3$$ etc.
Result: Formation of an alkyl fluoride $$R\!-\!F$$ (List II-IV).

Case C - Hunsdiecker reaction
Definition: Decarboxylative bromination of a silver salt of a carboxylic acid ( $$RCOOAg$$ ) with $$Br_2$$ in $$CCl_4$$.
Result: An alkyl bromide with one carbon atom fewer, $$R\!-\!Br$$ (List II-I).

Case D - Sandmeyer reaction
Definition: Substitution of the diazonium group in an aromatic diazonium salt by $$Cl$$, $$Br$$ or $$CN$$ using cuprous salts ( $$CuCl,\; CuBr,\; CuCN$$ ).
Result: Aryl chloride, aryl bromide or benzonitrile (List II-II).

Collecting the correct matches:
A → III (Finkelstein)
B → IV (Swarts)
C → I (Hunsdiecker)
D → II (Sandmeyer)

Therefore, the correct combination is:
Option B which is: A-III, B-IV, C-I, D-II.